用相同的参数调用基类的C ++构造函数

Question

I've ve something like this:

class A {
public:
  A(int a1,int a2,int a3) {
  }
}
class B: public A {
public:
  B(int a1,int a2,int a3) : A(a1,a2,a3) { 
     .. Do Some More Init for B ...
  }
}

Is it possible to write the constructor in aa shorter form? The Constructor of the base class should be called with the same arguments. But I don't want to list all of the arguments explicit. And I need to do some special initialization stuff for class B.

I think something like this:

class B: public A {
public:
  B(int a1,int a2,int a3) : A(...AUTOARG...) { 
    .. Do Some More Init for B ...
  }
}

Is this possible in C++/11?

Answer 1

If there is no code happening inside B 's constructor and you simply want to allow its construction in the same ways A is constructible, you can inherit A 's constructors:

class B: public A {
public:
  using A::A;
};

This will give B all the same constructors as A has.

If, as the edited question says, B 's constructor is not empty, you can still get "do all A does plus some more" in a generic (but not as nice) way using a "catch-all" constructor with perfect forwarding:

class B: public A {
public:
  template <class... Arg>
  B(Arg&&... arg) : A(std::forward<Arg>(arg)...) { 
     //.. Do Some More Init for B ...
  }
}

Note that this changes B 's behaviour slightly with respect to type traits. It now seems constructible from anything, but using incorrect arguments will fail to compile. If you also need the class to play nice with constructibility detection traits etc., you can conditionally disable the incorrect instantiations using SFINAE .

Answer 2

Better still, inherit the constructor. This is a new feature from C++11:

class B: public A {
public:
  using A::A;
};

Unfortunately though, if you want the B constructor to do additional things, then you'll need to write it out longhand as you currently do.